Characterization of Derivability from a Limit Condition

Question

Problem: $f:R\rightarrow R$ is continuous at $0$. If for all $a_n$ and $b_n$ satisfying $a_n\neq b_n\neq 0$ and $\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n=0$ there is a unique real number $k$ such that $$\lim_{n \to \infty} \frac{f(b_n)-f(a_n)}{b_n-a_n}=k$$Prove or give a counterexample: $f$ is derivable at $0$ and $f'(0)=k$.
Note: The problem arises from proving its converse that is correct: if $f$ is derivable at $0$ and $a_n$ and $b_n$ are defined as before, then $\lim_{n \to \infty} \frac{f(b_n)-f(a_n)}{b_n-a_n}=f'(0)$.
My approach is define $g(h)=\frac{f(h)-f(0)}{h}$, then put $a_n, b_n$ into $h$ to get $$k=\lim_{n \to \infty} \frac{f(b_n)-f(a_n)}{b_n-a_n}=\lim_{n \to \infty}{\frac{b_ng(b_n)-a_ng(a_n)}{b_n-a_n}}$$But I'm not sure whether to take the limit to get $g(0)=k$.
Additional Question: What if $f$ is not necessarily continuous at $0$?

Answer 1

The result is true. We can take any arbitrary sequence $a_n$ with $a_n\neq 0$ and $a_n\to 0$ and a fixed number $c\neq 1$ and put $b_n=ca_n$ so that $b_n\neq a_n$ and $b_n\neq 0$ and $b_n\to 0$.

Then by hypothesis $$\lim_{n\to \infty} \frac{f(ca_n) - f(a_n)} {(c-1)a_n}=k$$ Since $a_n$ is arbitrary this means that $$\lim_{x\to 0}\frac{f(cx)-f(x)}{x}=k(c-1)$$ And using a tricky argument given in this answer one shows that $f'(0)$ exists and equals $k$. The continuity of $f$ at $0$ is necessary for the result to hold.